Question

A carbon dioxide laser emits sinusoidal electro-magnetic wave that travels in vacuum in the negative $x$ -direction. The wavelength is $10.6\mu m$ and $\overrightarrow{E}$ fields is parallel to z-axis, with $E_{max}=1.5\times 10^{6}Mv{m}^{-1}$ Then vector equations for $\overrightarrow{E}$ and $\overrightarrow{B}$ as a function of time and position are:-

• A. $\overrightarrow{E}=\hat{k}\left[1.5\times {\left(10\right)}^{6}cos\left(8.93\times {\left(10\right)}^{5}x+3.78\times {\left(10\right)}^{14}t\right)\right]v{m}^{-1}$
  $\overrightarrow{B}=\hat{j}\left[5.0\times {\left(10\right)}^{-3}cos\left(8.93\times {\left(10\right)}^{5}x+3.78\times {\left(10\right)}^{14}t\right)\right]T$
• B. $\overrightarrow{E}=\hat{k}\left[1.5\times {\left(10\right)}^{6}cos\left(8.93\times {\left(10\right)}^{5}x+3.78\times {\left(10\right)}^{14}t\right)\right]v{m}^{-1}$
  $\overrightarrow{B}=-\hat{j}\left[5.0\times {\left(10\right)}^{-3}cos\left(8.93\times {\left(10\right)}^{5}x+3.78\times {\left(10\right)}^{14}t\right)\right]T$
• C. $\overrightarrow{E}=\hat{k}\left[1.5\times {\left(10\right)}^{6}cos\left(5.93\times {\left(10\right)}^{5}x+1.78\times {\left(10\right)}^{14}t\right)\right]v{m}^{-1}$
  $\overrightarrow{B}=-\hat{j}\left[5.0\times {\left(10\right)}^{-3}cos\left(5.93\times {\left(10\right)}^{5}x+1.78\times {\left(10\right)}^{14}t\right)\right]T$
• D. $\overrightarrow{E}=\hat{k}\left[1.5\times {\left(10\right)}^{6}cos\left(5.93\times {\left(10\right)}^{5}x+1.78\times {\left(10\right)}^{14}t\right)\right]v{m}^{-1}$
  $\overrightarrow{B}=\hat{j}\left[5.0\times {\left(10\right)}^{-3}cos\left(5.93\times {\left(10\right)}^{5}x+1.78\times {\left(10\right)}^{14}t\right)\right]$

Answer 1

Answer: (D)

Let $I={\lim }_{n\to \infty }{\sum }_{r=1}^{4n}\frac{\sqrt{n}}{\sqrt{r}(3\sqrt{r}+4\sqrt{n}}$
$={\lim }_{x\to \infty }{\sum }_{r=1}^{4m}\frac{1}{n}\cdot \frac{1}{\sqrt{\frac{\Gamma }{n}}}\cdot \frac{1}{\frac{(3\sqrt{r}+4\sqrt{n}}{n}}$
$={\lim }_{n\to \infty }{\sum }_{r=1}^{4n}\frac{1}{\sqrt{\frac{\Gamma }{n}}}\cdot \frac{1}{(3\sqrt{\frac{\Gamma }{n}}+4}\cdot \frac{1}{n}$
$={\int }_0^4\frac{1}{\sqrt{x}(3\sqrt{x}+4}dx$
Put $3\sqrt{x}+4=t$
$\Rightarrow 3\cdot \frac{1}{2\sqrt{x}}dx=dt$
$\Rightarrow \frac{1}{\sqrt{x}}dx=\frac{2}{3}dt$
$\Rightarrow I={\int }_4^{10}\frac{1}{t^2}\cdot \frac{2}{3}dt=-\frac{2}{3}{\left[\frac{1}{t}\right]}_t^{10}$
$\Rightarrow a=\frac{2}{3}\left(\frac{1}{10}-\frac{1}{4}\right)=\frac{1}{10}$
$\Rightarrow a=10$